$g(t) = 2t^{2}-t+5(f(t))$ $f(t) = 3t+1$ $h(n) = -6n^{2}+5n+6-3(f(n))$ $ h(f(2)) = {?} $
First, let's solve for the value of the inner function, $f(2)$ . Then we'll know what to plug into the outer function. $f(2) = (3)(2)+1$ $f(2) = 7$ Now we know that $f(2) = 7$ . Let's solve for $h(f(2))$ , which is $h(7)$ $h(7) = -6(7^{2})+(5)(7)+6-3(f(7))$ To solve for the value of $h$ , we need to solve for the value of $f(7)$ $f(7) = (3)(7)+1$ $f(7) = 22$ That means $h(7) = -6(7^{2})+(5)(7)+6+(-3)(22)$ $h(7) = -319$